Question

Write the value of 2 (sin⁶ θ + cos⁶ θ) −3 (sin⁴ θ + cos⁴ θ) + 1.

Answer 1

$$\begin{gathered} 2\left({\sin }^6\theta +{\cos }^6\theta \right)-3\left({\sin }^4\theta +{\cos }^4\theta \right)+1 \\ =2\left({\sin }^2\theta +{\cos }^2\theta \right)\left({\sin }^4\theta +{\cos }^4\theta -{\sin }^2\theta .c{\mathrm{os}}^2\theta \right)-3\left({\sin }^4\theta +{\cos }^4\theta \right)+1 \\ =2.1\left({\sin }^4\theta +{\cos }^4\theta -{\sin }^2\theta .c{\mathrm{os}}^2\theta \right)-3\left({\sin }^4\theta +{\cos }^4\theta \right)+1 \\ =2\left({\sin }^4\theta +{\cos }^4\theta \right)-2{\sin }^2\theta .c{\mathrm{os}}^2\theta -3\left({\sin }^4\theta +{\cos }^4\theta \right)+1 \\ =-\left({\sin }^4\theta +{\cos }^4\theta \right)-2{\sin }^2\theta .c{\mathrm{os}}^2\theta +1 \\ =-\left\{{\sin }^4\theta +{\cos }^4\theta +2{\sin }^2\theta .c{\mathrm{os}}^2\theta \right\}+1 \\ =-{\left({\sin }^2\theta +{\cos }^2\theta \right)}^2+1 \\ =-1+1 \\ =0 \end{gathered}$$